Happy Valentine's Day, once again

As Brian explains, we have the two equations (for quantity and price)

$5d+7t+9c+3r=100$

$3d+5t+7c+9r=100$

The reduced row-echelon form of this system comes out to be

$d=c+12r-50$ and $t=50-2c-9r$

To end up with a some daisies and tulips in the bunch, we need $c+12r>50$ and $2c+9r<50$

The "final filtering" isn't so tedious now, since we deal with only two variables and there are no additional divisibility conditions to satisfy.

The second inequality gives $r\leq 5$ .

If $r=5$ then $2c<50-45$ so $c\leq 2$ , which gives two solutions.

If $r=4$ then $2<c<7$ , which gives four solutions.

If $r\leq 3$ , then there are no solutions.

Thus we have $\boxed{6}$ solutions overall.

Let $d,t,c,r$ be the respective number of bunches of daisies, tulips, carnations and roses.

The "quantity equation" is then $5d + 7t + 9c + 3r = 100$ , and the "price equation" is $3d + 5t + 7c + 9r = 100$ .

Subtracting the former from the latter yields the equation

$-2d - 2t - 2c + 6r = 0 \Longrightarrow 3r = d + t + c.$

Substituting this result back into the quantity equation yields

$6d + 8t + 10c = 100 \Longrightarrow 3d + 4t + 5c = 50$ ,

where $d,t,c$ must be positive integers and their sum must be divisible by $3$ in order for $r$ to be a positive integer as well. So now the tedious part, working our way down from $c = 8$ , (clearly $c = 10,9$ will not yield solutions), to $c = 1$ , checking for suitable values for $d,t$ and then the divisibility condition. This leads to the solution quadruples, where $r = \frac{d + t + c}{3}$ ,

$(d,t,c,r) = (4,2,6,4), (3,4,5,4), (2,6,4,4), (1,8,3,4), (12,1,2,5)$ and $(11,3,1,5)$ ,

i.e., a total of $\boxed{6}$ different ways.